So we can rule out the possibility of a "9" at r2c4, because there's either a "9" at r2c1, or else there's a "9" at r9c4. So the "9" in column 6 must lie in the top center 3x3 box, and we can also rule out a "9" at r8c6, making that cell into the {4, 6} pair.

It's definitely helpful, David. As you know, I'm looking for ways to make further progress on the puzzle, but more importantly, I'm looking to train myself on how to approach these things in general and how to spot the patterns that matter.

These are the hardest puzzles of the ones that I have tried on a regular basis. I don't know how the Sudoku community views the "Daily Nightmare", but I have considerably more success with the latter than with the "Tough" ones from the Australian site.

So we can eliminate the "5" at r9c4, creating the {1, 6, 9} triplet in column 4 and allowing us to make quite a few "easy" moves. It's not enough to solve the puzzle completely, but at least it gets the ball rolling.

In general I've found the "double-implication chain starting from pairs" technique to be very helpful with puzzles like this one. The one exception I've noticed is that the technique is hard to apply if there aren't very many pairs in the starting grid. dcb

PS I managed to finish this puzzle off with a couple more double-implication chains, all of them rooted in r5c2, or in the adjacent cell, r5c3. Let me know if you still had trouble with this puzzle -- I'll be glad to explain the rest of it, but it will probably be more fun if you finish it up yourself.

David, using your suggestions, I made quite a bit of progress, or so I thought, until I backed myself into a corner with three cells in c9 containing only the pair "39." Trying to learn techniques is a challenge, but accuracy has plagued me, even though I have always been good with details. I erased and started over.

Ravel, what are "empty rectangles"? That's a term I haven't heard before.

You have a strong link for 1 in r5c3-r5c6, the empty rectangle is in r12c12 of box 1 (in this case there are 2 ER's for 1 in box 1).
You can see, that either
r5c6=1 => r3c6<>1 or
r4c3=1 => r1c(12)3<>1 => r3c1(2)=1 => r3c6<>1

You have a uniqueness pattern here, which allows you to eliminate 6 from r1c3, ...

Not to put too fine a point on it, ravel, but I don't understand how "uniquity" applies in this situation -- the corners of the rectangle lie in four different houses. So you can't permute the values 1 & 6 around the corners, because that will produce an invalid grid (two ones in box 1 and box 5; two sixes in box 2 and box 4).

Interestingly, I did notice that setting r5c3 to "1" will lead to a contradiction when I first started working on this puzzle. The problem is, the path to the contradiction was so long and involved (at least, the one I found was) that it looked too much like "trial and error" to be any fun.

ravel wrote:

... the empty rectangle is in r12c12 of box 1 (in this case there are 2 ER's for 1 in box 1).

I read Havard's post about "empty rectangles", and I have to admit it didn't really grab my imagination. For me, at least, most of these "ERs" are more easily understood as double-implication chains.

This is exactly the same logic as the "empty rectangle" you described, ravel. I've got nothing against coining new terminology, but I'm not real certain what this particular idea adds. It may make it easier to spot cases where a DIC can be employed ... I don't know. dcb

... I don't understand how "uniquity" applies in this situation -- the corners of the rectangle lie in four different houses.

Darned, you are very right, i made the same mistake, that i have pointed out at least 2 times in other posts Seems that i was too anxious for finding uniqueness patterns recently.
Thanks for correcting that.

Quote:

This is exactly the same logic as the "empty rectangle" you described, ravel. I've got nothing against coining new terminology, but I'm not real certain what this particular idea adds.

Your double-implication chain describes the use of the ER in box 2, which - combined with the same strong link - leads to the elimination of 1 in r1c3.
Of course you are right, that each ER elimination can be notated as such a chain (by definition), but note that the ER would also work, if you had more than 2 candidates in r5c36 and an additional 1 in r1c4, which makes the chain a bit less easy to spot and a bit more complicated.
The ER would be spotted as easy as in this case: when you have a strong link, just look up/down (or right/left), if there is an ER.
So i would compare it to an xy-wing, which also commonly is considered to be an own technique, while its nothing but a simple double-implication chain.

Interestingly, I did notice that setting r5c3 to "1" will lead to a contradiction when I first started working on this puzzle. The problem is, the path to the contradiction was so long and involved (at least, the one I found was) that it looked too much like "trial and error" to be any fun.

David, I've spoken about trial-and-error in at least one other thread, perhaps responding to comments by Alan. If I've interpreted the remarks correctly, it seems the label "trial-and-error" is being applied to the result, not the method.

If you start out testing each of two values in a cell and find some other cells are the same for both values in the starting cell, then you've done a nice DIC or forcing chain. But if one value leads to a contradiction, then it's viewed negatively as trial-and-error. In my unsophisticated newbie world, the negatively viewed latter forces a value in the starting cell, while the positively viewed former forces value(s) in other cells.

I'm having trouble understanding the difference in attitude between the two.

But if one value leads to a contradiction, then it's viewed negatively as trial-and-error.

It's not the contradiction I object to, Marty -- it's the length of the "forcing chain." Here, let me illustrate the point. The puzzle that started this thread can be reduced to this state by doing some coloring, etc.

Clearly the grid cannot now be completed, because we have a "6" at r1c6, and the pair {2, 6} in r2c1 & r8c1. This is a contradiction, from which we conclude that r5c3 = 6.

I only had two reasons for avoiding this approach and seeking a simpler one when I worked up the previous solution (using shorter chains from r5c2).

-- This forcing chain is so long and involved that it's hard to explain.
-- Even after setting r5c3 = 6 one can't solve the puzzle very easily. Later on you'll have to find another forcing chain to finish it.

Generally I prefer shorter, more direct chains to longer, more complex ones. So I chose to seek another route to the solution to this puzzle. I don't have any objections to proofs by contradiction -- I understand that the statements "A is true" and "(not A) is false" are logically equivalent, and I always treat them that way. dcb

I agree David, some of the chains are extrememely long and I always wonder if I made a mistake, since I continue to have a problem with accuracy. On some of those chains, one of the values occasionally solves every remaining cell.

In the meantime, I'm not sure where logic ends and trial-and-error starts.